Question

A 19.247 g sample of petrified wood was found in a petrified forest. The sample showed a 14C activity of 12.9 decays/min. How long has the tree been dead? The halflife of carbon-14 is 5730 years and freshly cut wood contains 6.5 × 1010 atoms/g of 14C and there are 365.25 days in a year. Answer in units of y.

Answer 1

N = N0*exp(- lambda*t)

where lambda = 0.693 / Half-Life

lambda = 0.693/5730 =1.209*10^-4 per year = 3.835 *10^-12 per second

The 19.247 g sample has an activity 12.9decays/min = 12.9/60 decays/s = 0.215 Bq
So 1g of the sample has an activity A = 0.215/19.247 = 0.01117 Bq

The number, N, of C14 atoms in 1g of the sample is found using:
A = λN
0.01117 = 3.835 *10^-12 N
N =2.912x10⁹ C14 atoms/g

Freshly cut wood has a value N₀ = 6.5x10¹⁰ C14 atoms/g; this has dropped to 2.912x10⁹in time t
N = N₀e^(-λt)
2.912x10⁹= 6.5x10¹⁰ e^(-3.835 *10^-12t)
0.04481 = e^(-3.835 x 10⁻¹²t)
ln(0.04481) = -3.835 x 10⁻¹²t
-3.1052 = = -3.835 x 10⁻¹²t
t = 8.09x10¹¹ s
= 8.09x10¹¹ / (365 x 24 x 3600)
t= 25676 years